Math 4124 Wednesday, February 1 Second Homework Solutions 1. 2.1.10(a) on page 48 Prove that if H and K are subgroups of the group G , then so is H ∩ K . Let e be the identity of G . Then e ∈ H , K because H and K are subgroups of G , consequently e ∈ H ∩ K . Next let x , y ∈ H ∩ K . Then xy ∈ H because H ≤ G and xy ∈ K because K ≤ G . Therefore xy ∈ H ∩ K . Finally x-1 ∈ H and K because H and K are subgroups of G , and we deduce that x-1 ∈ H ∩ K . It now follows that H ∩ K is a subgroup of G as required. 2. 1.2.3 on page 27 Use the standard generators and relations to show that every ele-ment of D 2 n which is not a power of r has order 2. Deduce that D 2 n is generated by the two elements s and sr , both of which have order 2. Each element of D 2 n can be written uniquely in the form r i or sr i , where 0 ≤ i ≤ n-1 (see page 25). The elements r i are of course powers of r , so we need to prove that sr i has order 2. Since sr i ± = e , it will be sufﬁcient to show that ( sr i ) 2 = e ; we shall show that this is true for all i . We have ( sr i ) 2 = sr i sr i . Now the relation rs = sr-1 applied i times shows that r i s = sr-i , consequently ( sr i ) 2 = ssr-i r i = s 2 r-i + i = ee = e , which is what is required. Finally we need to show that D 2 n is generated by the two elements s and sr , both of which have order two. That they both have order two follows from the previous paragraph. Also every element of D 2 n can be written as a product of r , s , r-1 , s-1 , and since r = s-1 ( sr ) , we see that every element can be written as a product of s , ( sr ) , s-1 , ( sr )-1 . This establishes that D 2 n is generated by s and sr .

This is the end of the preview. Sign up
to
access the rest of the document.